Let $\{a_n\}$ be a sequence. If $a,b$ given so that $\lim\limits_{n\rightarrow\infty} a_n=a$ and $\lim\limits_{n\rightarrow\infty} a_n=b$, then $a=b$.


Let $\epsilon>0$. Since $\lim\limits_{n\rightarrow\infty} a_n=a$ and $\lim\limits_{n\rightarrow\infty} a_n=b$, we can have $\vert a_n-a\vert<\epsilon$ and $\vert a_n-b\vert<\epsilon$.

Then add $\vert a_n-a\vert<\epsilon$ and $\vert a_n-b\vert<\epsilon$, we have $$\vert a_n-a\vert+\vert a_n-b\vert<2\epsilon$$ and this is equivalent to $$\vert a_n-a\vert+\vert b-a_n\vert<2\epsilon$$ Applying the triangle inequality, we can have $$\vert (a_n-a)\vert-\vert(b-a_n)\vert\leq\vert (a_n-a)+(b-a_n)\vert\leq\vert a_n-a\vert+\vert b-a_n\vert$$ $$\Longrightarrow\vert (a_n-a)\vert-\vert(b-a_n)\vert\leq\vert b-a\vert\leq\vert a_n-a\vert+\vert b-a_n\vert\tag1$$

so $0\leq\vert b-a\vert$ and implies $a=b\tag2$

From $(1)$ to $(2)$, I am not sure correct or not. But it seems right to me because the greater equal site can imply $0=\vert b-a\vert$.

Can anyone check my solution? If that is not right, please give me a hit or suggestion to write a better one. (I seen a solution by using contraction)

  • 1
    $\begingroup$ Euclid (circa 2500 years ago) : Things equal to the same thing are equal to each other. If f=a and f=b then it does not matter how complicated f is, it's a thing. And the things a,b are equal to the same thing, that is, to f, so a=b.Period. $\endgroup$ – DanielWainfleet Oct 11 '15 at 3:57
  • $\begingroup$ @user254665: Why they give proof for this! I cannot understand! $\endgroup$ – H. R. Oct 11 '15 at 20:10
  • $\begingroup$ I suspect this is a school problem to see which students are even reading the assignment. I know that some teachers will sometimes include a "free" question, but you won't get the marks for it if you don't read it. $\endgroup$ – DanielWainfleet Oct 11 '15 at 22:16

Very nice start to the proof! There are some things a little shady with the proof, but the most important point I want to make is this:

When you are writing proofs, you should try to tell the reader a story. When I read your proof, I feel like you are trying to conform to some type of proof template you have in your mind. Proofs don't have to follow a specific structure or style -- you can use a lot of words in your proof! The only thing that must be true about your proof is that each step logically follows from the previous one. With that said, let me tell you how I would have written the proof of this -- I am writing this proof off of the top of my head because my goal is just to tell you a story of how I understand each step (that's what a proof should do -- tell the reader how you understand each step).


I want to prove if $a$ and $b$ are real numbers, and $a_{n}$ is a sequence such that $\lim \limits_{n \to \infty} a_{n} = a$ and $\lim \limits_{n \to \infty} a_{n} = b$, then $a = b$.

Well, to show that $a = b$, I can show that $|a - b| = 0$, since if $x$ is a real number, then $|x| = 0$ is equivalent to $x = 0$. So, I will show that $|a-b| = 0$ (which means I need to start playing around with $|a-b|$ to show this).

But for each natural number $n$, $|a - b| = |a - a_{n} + a_{n} - b| \leq |a - a_{n}| + |a_{n} - b|$ (by the triangle inequality).

But now I know $\lim \limits_{n \to \infty} a_{n} = a$, so that means for any $\epsilon > 0$, I can find some $N$ such that $n \geq N$ implies $|a - a_{n}| < \frac{\epsilon}{2}$.

Similarly, since I know $\lim \limits_{n \to \infty} a_{n} = b$, that means I can $N'$ such that $n \geq N'$ implies $|a_{n} - b| < \frac{\epsilon}{2}$.

Then, if I take $M = \max \{N, N'\}$, then if $n \geq M$, we have both $|a_{n} - a| < \frac{\epsilon}{2}$ and $|b -a_{n}| < \frac{\epsilon}{2}$.

Since I showed above $|a - b| \leq |a - a_{n}| + |a_{n} - b|$ for every $n \in \Bbb N$, that means the inequality is certainly true if $n \geq M$. But if $n \geq M$, then the inequality implies $|a - b| < \epsilon$. But $\epsilon$ was arbitrary! So, since $|a-b| < \epsilon$ for every $\epsilon > 0$, that means $|a - b| = 0$, so $a = b$.

  • $\begingroup$ I understand that $\epsilon$ is a arbitrary number that we can pick any number, but I don't quite follow that ''since$\vert a-b\vert<0...$ that means$\vert a-b\vert=0$ $\endgroup$ – Simple Oct 11 '15 at 3:10
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    $\begingroup$ @Simple Good question. I never mentioned the line $|a - b| < 0$ in the proof. But instead, I am using the fact that if $|a - b| < \epsilon$ for every positive number $\epsilon$, then $|a - b| = 0$. Why is this? Well, by the definition of absolute value, $|a-b| \geq 0$, right? Now, suppose $|a-b| < \epsilon$ for every $\epsilon > 0$. Suppose also that $|a - b| \neq 0$. Then $|a - b| > 0$, right? But then let $\epsilon ' = \frac{|a-b|}{2}$. Then $\epsilon '$ is a positive number and $|a - b| \not < \epsilon '$, which contradicts the assumption that $|a - b| < \epsilon$ for every $\endgroup$ – layman Oct 11 '15 at 3:13
  • $\begingroup$ @Simple ... $\epsilon > 0$. So, that means $|a - b|$ must $= 0$ if $|a - b| < \epsilon$ for every $\epsilon > 0$. $\endgroup$ – layman Oct 11 '15 at 3:13
  • $\begingroup$ Thanks, that is very helpful and clear $\endgroup$ – Simple Oct 11 '15 at 3:16
  • $\begingroup$ @Simple I am glad to help. :) $\endgroup$ – layman Oct 11 '15 at 3:16

Let $\epsilon =|a-b|$ and assume for contradiction that $\epsilon>0$.

Then there is some $N, N'$ such that for $n\geq N$ and $n\geq N'$, $|a_n-a|<\epsilon/2$ and $|a_n-b|<\epsilon/2$ respectively. Let $M=\max\{N,N'\}$. Then $|a_M-a|<\epsilon/2$ and $|a_M-b|<\epsilon/2$.

Then applying triangle inequality:


A contradiction, so $|a-b|=0$ and $a=b$.


A straightforward argument would be to show $|a-b|< \epsilon$ which is given below. I ignored the use of the index for which the sequence converges to $a,b$. We know to just take the max of the two $N_1,N_2$.

$|a_n- a|<\frac{\epsilon}{2}$ and $|a_n-b|< \frac{\epsilon}{2}$

$|a-b| = |a-a_n+a_n-b| \leq |a-a_n|+|a_n-b| < \epsilon \Rightarrow a =b$


"Let ϵ>0. Since lim n→∞an=a and limn→∞an=b, we can have |an−a|<ϵ and |an−b|<ϵ. "

Not quite. Not for any epsilon and any n. If for any specific epsilon, this is true for significantly large values of n. Not all n. In other words for every $\epsilon > 0$, there exists an N that is sufficiently large so that n > N means n will be sufficiently large for that to be true.

So better writing:

Let $\epsilon > 0$. Since $\lim\limits_{n\rightarrow\infty} a_n=a$ there exists an $N$ such that $n > N$ implies $\vert a_n-a\vert<\epsilon$ and there exists an $M$ such that $n > M$ implies $\vert a_n-b\vert<\epsilon$. So for $n > max(M,N)$:

$\vert a_n-a\vert+\vert a_n-b\vert<2\epsilon$ so

$\vert a - a_n\vert+\vert a_n-b\vert<2\epsilon$

$\vert a - b\vert <= \vert a - a_n\vert+\vert a_n-b\vert<2\epsilon$ (by triangle inequality.)

(From here there are many ways to show a must equal b. I think a simple contradiction is the clearest. It lets us practice epsilons.) If $a \ne b$ then $|a - b| > 0$. If we chose $0< \epsilon < |a - b|/2$ we get a contradictory result.

$|a - b| < 2\epsilon < 2|a -b|/2 = |a - b|$.

This is impossible so |a - b| = 0 and a = b.


Careful how you choose your $n$. Remember that the definition says "Given $\varepsilon>0$ there exists $N$ such that for all $n \geq N$ then $|a_n-a|<\varepsilon$". A similar index $M$ can be found for $|a_m-b|<\varepsilon$ to be true. Then take the largest of the two. We can assume without loss of generality that $N \geq M$, and in that way you are then safe to say $|a_n-b|<\varepsilon$ and $|a_n-a|<\varepsilon$ are simultaneously true for all $n \geq N$.

To streamline the proof, you can actually say $$\vert a_n-a\vert+\vert b-a_n\vert \geq \vert (a_n-a)+(b-a_n)\vert \\ = \vert b-a \vert $$ to see that $$\vert b-a \vert < 2\varepsilon$$ in a few less implications.

Next I can show you a little "trick" to eliminate that $2$ that appears attached to the $\varepsilon$. State that "given $\varepsilon>0$ where $\varepsilon ' = \frac{\varepsilon}{2}$, I can find indices $N',M'$ (dummy indices to go with $\varepsilon '$) such that $|a_n-a| < \varepsilon '$ and $|a_m-b| < \varepsilon '$ for all $n \geq N'$ and for all $m \geq M'$. WLOG take $N'$ to be larger again. Now when you add your inequalities you will just be left with $$\vert b-a \vert < \varepsilon$$ At this point you are done, because the inequality holds for all $n \geq N'$ and an arbitrary $\varepsilon$.


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